Summation of product of $m+1$ alternate numbers We know that 
$$\begin{align}
\sum_{r=1}^n \prod_{j=0}^m (r+j)&=\sum_{r=1}^n r(r+1)(r+2)\cdots(r+m)\\
&=(m+1)!\sum_{r=1}^n\binom {r+m}{m+1}\\
&=(m+1)!\binom {n+m+1}{m+2}\\
&=\frac {(n+m+1)^\underline{m+2}}{m+2}\end{align}$$
but is there a similar closed form for 
$$\sum_{r=1}^n \prod_{j=0}^m (r+2j)=\sum_{r=1}^n r(r+2)(r+4)\cdots (r+2m)$$
?

Edit 1
Here's a summary of results from wolframlpha for the first few values of $m$ (please excuse the messy alignment).

From the summary we observe some interesting patterns as follows:


*

*all results have factors $n,(n+1)$

*all results have a factor of $\frac 1{2(m+2)}$ or $\frac 1{m+2}$ where the product of the other factors results in a $2n^{m+2}$ term or $n^{m+2}$ term respectively

*for $m=2$: $(n+4),(n+5)$ are factors

*for $m=4$: $(n+8),(n+9)$ are factors

*for $m=6$: $(n+12),(n+13)$ are factors

*for $m=8$: $(n+16),(n+17)$ are factors

*this implies that for even $m$: $(n+2m), (n+2m+1)$ are factors


Perhaps the observations above might be helpful in working out the solution.
The solution could then possibly be of the form:
$$\begin{cases}
\dfrac {n(n+1)(n+2m)(n+2m+1)\cdot f(n,m)}{2(m+2)}\qquad \text{for even $m$}\\
\dfrac {n(n+1)\cdot g(n,m)}{2(m+2)}\qquad \qquad\qquad\qquad\qquad\text {for odd $m$}
\end{cases}$$ 
 A: \begin{align*}\sum_{r=1}^n r(r+2)(r+4)...(r+2m)&=\sum_{1\le r=2k\le n}r(r+2)(r+4)...(r+2m)\\&+\sum_{1\le r=2k+1\le n}r(r+2)(r+4)...(r+2m) \\ &=\underbrace{\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{(2k+2m)!!}{(2k-2)!!}}_{S_1}+\underbrace{\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\frac{(2k+2m+1)!!}{(2k-1)!!}}_{S_2}\end{align*}
where: $\quad (2x)!!=2^x .x!\quad $ and $\quad (2x-1)!!=\dfrac{(2x)!}{(2x)!!}$
Note that: $\quad 0!!=2^0.0!=\frac{0!}{0!!}=(-1)!!=1$
\begin{align*} S_1&=(2m+2)!!+\frac{1}{2m+4}\sum_{k=2}^{\left\lfloor\frac{n}{2}\right\rfloor}\Delta\left(\frac{(2m+2k)!!}{(2k-4)!!}\right)\\ &=(2m+2)!!+\frac{1}{2m+4}\cdot\frac{\left(2m+2\left\lfloor\frac{n}{2}\right\rfloor+2\right)!!}{\left(2\left\lfloor\frac{n}{2}\right\rfloor-2\right)!!}-\frac{1}{2m+4}\cdot\frac{(2m+4)!!}{0!!}\\&=\frac{1}{2m+4}\cdot\frac{\left(2m+2\left\lfloor\frac{n}{2}\right\rfloor+2\right)!!}{\left(2\left\lfloor\frac{n}{2}\right\rfloor-2\right)!!}\end{align*}
\begin{align*} S_2&=(2m+1)!!+\frac{1}{2m+4}\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\Delta\left(\frac{(2m+2k+1)!!}{(2k-3)!!}\right)\\ &=(2m+1)!!+\frac{1}{2m+4}\cdot\frac{\left(2m+2\left\lfloor\frac{n+1}{2}\right\rfloor+1\right)!!}{\left(2\left\lfloor\frac{n+1}{2}\right\rfloor-3\right)!!}-\frac{1}{2m+4}\cdot\frac{(2m+3)!!}{(-1)!!}\\&=\frac{(2m+1)!!}{2m+4}+\frac{1}{2m+4}\cdot\frac{\left(2m+2\left\lfloor\frac{n+1}{2}\right\rfloor+1\right)!!}{\left(2\left\lfloor\frac{n+1}{2}\right\rfloor-3\right)!!}\end{align*}
A: Simply for my result, I have:
$$\sum_{r=1}^n\prod_{j=0}^m(r+2j)=S(n,m)=\frac{1}{2m+4}\left(\prod_{k=0}^m(2k+1)+\prod_{k=0}^{m+1}(n+2k)+\prod_{k=0}^{m+1}(n+2k-1)\right)$$
That's a polynomial of $n$ with $m+2$ degrees.
The first few values of $m$
\begin{align*}S(n,1)&=\frac{1}{6}\left(\prod_{k=0}^1(2k+1)+\prod_{k=0}^2(n+2k)+\prod_{k=0}^2(n+2k-1)\right) \\ &=\frac{1}{6}\bigg(1.3+n(n+2)(n+4)+(n-1)(n+1)(n+3)\bigg)\\ &=\frac{1}{6}n(n+1)(2n+7)\end{align*}
\begin{align*}S(n,2)&=\frac{1}{8}\bigg(1.3.5+n(n+2)(n+4)(n+6)+(n-1)(n+1)(n+3)(n+5)\bigg) \\ &=\frac{1}{8}\cdot2n(n+1)(n+4)(n+5)\end{align*}
etc..,
I think so don't closed form for this sum, It is a similarity with the sum $\displaystyle \sum_{k=1}^n k^r$
